A wire mesh consisting of very small squares is viewed at a distance of `8cm` through a magnifying converging lens of focal length `10 cm`, kept close to the eye. The magnification produced by the lens is:

Options

A. `5`
B. `8`
C. `10`
D. `20`

Answer

a

Explanation

Lens formula is given by
`(1)/(f)=(1)/(v)-(1)/(u)` ...(i)
where `f` is focal length of lens, `v` is image distance and `u` is object distance.
Given, `f=10 cm` (as lens is converging)
`u= -8cm` (as object is placed on left side of the lens)
Substituting these values in Eq.(i), we get,
`(1)/(10)=(1)/(v)-(1)/(-8)implies (1)/(v)=(1)/(10)-(1)/(8)`
`(1)/(v)= (8-10)/(80)implies v=(80)/(-2)= -40 cm`
Hence, magnification produced by the lens
`m=(v)/(u)= (-40)/(-8)=5`